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Help with Laplace Transformations and 2nd order ODEs

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sara_87
#55
Nov26-08, 01:00 PM
P: 774
For 1/(s+1)^3, let's just look at 1/s^3 ; we know that t^2 is 2/s^3 so 1/s^3 will be the inverse of (1/2)(t^2) right?
but, we dont want 1/s^3, we want 1/(s+1)^3, so what do we do (like we did before)?
TFM
#56
Nov26-08, 01:13 PM
P: 1,031
First Shft Theorem

Ok, the first shift theorem says that:
if you have a function that looks like: e^(at)F(t);
the laplace of that is the laplace of F(t) but instead of putting p, you put p-a
eg: L(e^3t(t))
we know that laplace of t is 1/p^2 (you know that from the tables right?)
and so here a=3 so the laplace of e^(3t)(t)=1/(p-3)^2
so, where ever you have p, you replace with p-a in this case, p-3
Do you agree??
[tex] t^2 = 2/s^3 [/tex]

[tex] \frac{1}{2}t^2 = \frac{1}{s^3} [/tex]

using FST

[tex] \frac{1}{2}t^2 = \frac{1}{b^3} [/tex], where b = s + 1

would this make it:

\frac{1}{2}(t+1)^2 = \frac{1}{b^3} [/tex]

???

TFM
sara_87
#57
Nov26-08, 03:54 PM
P: 774
I dont see any e's in your answer (there should be).

now as we did before, the second fraction is the laplace transform of: 4te^(-2t)
because if you find the laplace of 4t, this will give you 4/s^2 but we dont want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t).

now, can you do the same for the 1/(s+1)^3
TFM
#58
Nov27-08, 08:37 AM
P: 1,031
Okay so:

4t will give you 4/s^2,but we don't want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t).
so would this mean:

[tex] t^2 would give: \frac{2}{s^3} [/tex]

thus would:

[tex] \frac{2}{(s + 1)^3} be \frac{1}{2}t^2 e^{-t} [/tex]

???

TFM
sara_87
#59
Nov27-08, 02:42 PM
P: 774
very good!!
TFM
#60
Nov27-08, 02:51 PM
P: 1,031
Great

So now:

[tex] Y = \frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2} [/tex]

this goes to:

[tex] Y = \frac{1}{2}t^2 e^{-t} + 4te^{-2t} [/tex]

So now we put this value of Y back into:

[tex] s^2(Y) - 4 + 4(sY) + 4Y = \frac{1}{s + 2} [/tex]

???

TFM
sara_87
#61
Nov27-08, 03:02 PM
P: 774
y(t)=(1/2)t^2(e^-t) + 4t(e^-2t)
This is the answer. dont substitute anything into anything.
TFM
#62
Nov27-08, 03:06 PM
P: 1,031
Excellent.

So

[tex] y(t) = \frac{1}{2}t^2 e^{-t} + 4te^{-2t} [/tex]

Tomorrow, I'll start (c), if thats okay?

TFM
sara_87
#63
Nov28-08, 03:45 PM
P: 774
yep thats fine
TFM
#64
Nov29-08, 12:12 PM
P: 1,031
Okay, so:

(c):

[tex] y'' + y = sin(t), y_0 = 1, y_0' = 0 [/tex]

L(y(t)) = Y
L(y'(t)) = sL(y(t)) - y(0)
L(y''(t)) = sL(y'(t))-y'(0) = s^2(L(y(t))-sy(0)-y'(0)


there is no y',


[tex] L(y(t)) = Y [/tex]
[tex] L(y'(t)) = sL(y(t)) - 1 [/tex]
[tex] L(y''(t)) = sL(y'(t))-0 = s^2(L(y(t))-sy(0)-y'(0) [/tex]


and for the sin(t)

[tex] sin(\alpha t) = \frac{\alpha}{p^2 + \alpha^2} [/tex]

since alpha = 1:

[tex] sin(t) = \frac{1}{p^2 + 1} [/tex]

Is this okay, I am not quite sure because the lack of y'

TFM
sara_87
#65
Nov29-08, 12:19 PM
P: 774
That is absolutely right.
what next?
TFM
#66
Nov29-08, 12:30 PM
P: 1,031
Okay so:

[tex] L(y(t)) = Y [/tex]

[tex] L(y'(t)) = sL(y(t)) - 1 [/tex]

[tex] L(y''(t)) = s^2(L(y(t))-sy(0) [/tex]

So now I substitute these into the original equation:

[tex] y'' + y = sin(t) [/tex]

[tex] s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1} [/tex]

Is this okay? Do I substitute y(0) into sy(0), or is there another value/leave it alone?

TFM
sara_87
#67
Nov29-08, 12:49 PM
P: 774
That's fine
substitute y(0) and also substitue the value L(y(t))
TFM
#68
Nov29-08, 12:59 PM
P: 1,031
Okay, so

[tex] s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1} [/tex]

[tex] s^2(Y)-1 + Y = \frac{1}{p^2 + 1} [/tex]

[tex] Ys^2 + Y - 1 = \frac{1}{p^2 + 1} [/tex]

Now I need to make the Y the subject:

[tex] Ys^2 + Y = \frac{1}{p^2 + 1} + 1 [/tex]

factorise out

[tex] Y(s^2 + 1) = \frac{1}{p^2 + 1} + 1 [/tex]

divide through:

[tex] Y = \frac{\frac{1}{p^2 + 1} + 1}{s^2 + 1} [/tex]

Now I need to find the inverse.

Firstly, split it up into two fractions:

[tex] Y = \frac{\frac{1}{p^2 + 1}}{s^2 + 1} + \frac{1}{s^2 + 1} [/tex]

Does this look okay?

TFM
sara_87
#69
Nov29-08, 01:42 PM
P: 774
in ur second line, you have: s^y-1+y=1/(s^2+1)
the -1 on the left hand side should be -s (since you have:L(y'')=s(sy-1)
and remember: s is p so dont put both...choose one of them :)
TFM
#70
Nov29-08, 01:52 PM
P: 1,031
Okay so:

[tex] s^2(Y)-1 + Y = \frac{1}{p^2 + 1} [/tex]

[tex] Ys^2 + sY - 1 = \frac{1}{s^2 + 1} [/tex]

Does this look okay now?

TFM
sara_87
#71
Nov29-08, 03:35 PM
P: 774
you should have:
Ys^2 - s + Y=1/(s^2+1)

i think you made a mistake while substituting.
TFM
#72
Nov29-08, 03:58 PM
P: 1,031
Okay, back a bit originally:

[tex] s^2(L(y(t))-sy(0) + Y = \frac{1}{s^2 + 1} [/tex]

y(0) = 1

L(y(t)) = Y

put these in:

[tex] s^2(Y) - 1 + Y = \frac{1}{s^2 + 1} [/tex]

goes to:

[tex] Ys^2 - 1 + Y = \frac{1}{s^2 + 1} [/tex]

I still have a minus one where there should be a minus s.????

TFM


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